# Sum of Cubes

How to solve the given summation by using the sum of cubes (k^{3}) formula: formula, 1 example, and its solution.

## Formula

### Formula

The sum of k^{3}, as k goes from 1 to n, is

[n(n + 1)/2]^{2}.

This means

1^{3} + 2^{3} + 3^{3} + ... + n^{3} = [n(n + 1)/2]^{2}.

## Example

### Example

### Solution

k(k^{2} + n) = (k^{3} + nk)

k^{3} is in the sigma.

See ∑ (+nk).

k is the variable of the sigma.

So think n as a coefficient.

Then n gets out from the sigma.

And k is in the sigma.

Summation: How to Solve

The sum of k^{3} is

[n(n + 1)/2]^{2}.

So [n(n + 1)/2]^{2} + n⋅n(n + 1)/2.

[n(n + 1)/2]^{2}

= n^{2}(n + 1)^{2}/2^{2}

= n^{2}(n + 1)^{2}/4

Power of a Product

Power of a Quotient

+n⋅n(n + 1)/2 = +n^{2}(n + 1)/2

To combine these two fractions,

change +n^{2}(n + 1)/2 to +2n^{2}(n + 1)/4.

The common factor of

n^{2}(n + 1)^{2}/4 and +2n^{2}(n + 1)/4

is n^{2}(n + 1)/4.

So write

n^{2}(n + 1)/4.

Write, n^{2}(n + 1)^{2}/4 ÷ n^{2}(n + 1)/4,

[(n + 1).

And write, +2n^{2}(n + 1)/4 ÷ n^{2}(n + 1)/4,

+2].

[(n + 1) + 2] = (n + 3)

So n(n + 1)(n + 3)/4 is the answer.